Question

Find the value of x, if the fourth term in the expansion of $(\frac{1}{x^2}+x^2{.2}^x{)}^6$ is $160$.

Answer 1

4th term $⟹T_{3+1}={6_C}_3\left(\frac{1}{x^2}\right).(x^2{)}^3.(2^x{)}^3$

Therefore, ${6_C}_3.(2^x{)}^3=160$

${20.2}^{3x}=160$

$2^{3x}=8$

$2^{3x}=2^3$

$3x=3$

$x=1$